Let $X,Y$ are nonsingular complex projective surfaces and $f:Y\to X$ be a birational morphism. Let $E_1,\ldots,E_k$ are irreducible exceptional divisors of $f$.

In this situation, we have $K_Y=f^{\ast}K_X+\sum_{i=1}^{k}m_iE_i (m_i>0)$ ($K_Y,K_X$: canonical divisor of $Y,X$)

How to prove $(f^{\ast}K_X)\cdot E_i=0 (\forall i)$?

More generally,Is it true that $(f^\ast D)\cdot E_i=0$ for all $i$ and for all divisor $D$ on $X$?

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    $\begingroup$ What are $f(E_i)$? Can you say that if $D$ is any divisor on $X$, it is linearly equivalent to a divisor disjoint from $f(E_i)$? $\endgroup$ – Mohan May 7 at 17:45

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